Question

Suppose \(f\) is a polynomial function. If \(f(-2)=7\) and \(f(6)=-1,\) then the Intermediate Value Theorem guarantees which of the following? Justify your answer. (a) \(f(0)=0\) (b) \(f(c)=3\) for at least one number \(c\) between -2 and 6 . (c) \(f(c)=0\) for at least one number between -1 and 7 . (d) \(-1 \leq f(x) \leq 7\) for all numbers in the closed interval [-2,6]

Short answer

Option (b) is correct because the Intermediate Value Theorem guarantees that there exists at least one \(c\) in \([-2, 6]\) such that \(f(c) = 3\).

Step-by-step solution

Understanding the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \(f\) is continuous on a closed interval \([a, b]\) and \(N\) is any number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \([a, b]\) such that \(f(c) = N\). A polynomial function is continuous everywhere.

Identification of Given Interval and Function Values

Given that \(f(-2) = 7\) and \(f(6) = -1\), the function \(f\) is continuous on the interval \([-2, 6]\). So we need to assess which of the given options can be guaranteed by the Intermediate Value Theorem.

Analysis of Option (a)

Option (a) suggests \(f(0) = 0\). To apply the Intermediate Value Theorem, \(0\) should be between the values \(7\) and \(-1\), and we need to check if these boundary values cross \(0\). However, we cannot make any conclusion regarding \(f(0)\) without additional information specific to the function \(f\). Therefore, option (a) is not guaranteed.

Analysis of Option (b)

Option (b) suggests \(f(c)=3\) for at least one number \(c\) between -2 and 6. Since \(3\) is between 7 and -1, the Intermediate Value Theorem guarantees that there exists at least one \(c\) in the interval \([-2, 6]\) such that \(f(c) = 3\). This option is guaranteed by the theorem.

Analysis of Option (c)

Option (c) suggests \(f(c)=0\) for at least one number between -1 and 7. The number \(c\) should be within the interval \([-2, 6]\). Since 0 is between the given values of \(f(-2) = 7\) and \(f(6) = -1\), the Intermediate Value Theorem guarantees that there is at least one \(c\) in the interval \([-2, 6]\) for which \(f(c) = 0\). However, this interval doesn't necessarily correspond to a function value between -1 and 7, making this option also feasible but less specific compared to option (b).

Analysis of Option (d)

Option (d) suggests that \(-1 \leq f(x) \leq 7\) for all numbers in the closed interval \([-2, 6]\). The Intermediate Value Theorem does not guarantee that the function \(f\) stays within any particular bounds except crossing all intermediate values between \(f(a) = 7\) and \(f(b) = -1\). Hence, option (d) is not guaranteed and could be excluding some values outside of this range.

Key concepts

Polynomial Function

A polynomial function is an expression of the form \[f(x) = a_n x^n + a_{n-1} x^{n-1} + \ \ + a_1 x + a_0\]\brwhere \(a_n, a_{n-1}, ..., a_1, a_0\) are constants, and \(n\) is a non-negative integer. One key trait of polynomial functions is that they are defined for all real numbers and can be graphed as a smooth, unbroken line or curve.

• No sharp corners
• No discontinuities
• No breaks

These functions follow predictable patterns and because of their smoothness, polynomial functions are inherently continuous. Understanding the nature of these functions is crucial when applying the Intermediate Value Theorem.

Continuous Function

A continuous function is where the output of the function changes smoothly without any jumps or gaps as the input changes. Mathematically speaking, for every point \(x = a\) within the domain:

• \(f(a)\) is defined
• The limit of \(f(x)\) as \(x\) approaches \(a\) exists
• The limit of \(f(x)\) as \(x\) approaches \(a\) is \(f(a)\)

By its characteristics, a polynomial function is continuous because it has no breaks or gaps in its graph. Understanding continuity is essential when using the Intermediate Value Theorem, which requires the function to be continuous over the interval being considered.

Value Guarantee on Interval

The Intermediate Value Theorem ensures that a continuous function will take on every value between \(f(a)\) and \(f(b)\) at least once within the interval \[a, b\]. In our exercise, the given function \(f(x)\) satisfies \(f(-2) = 7\) and \(f(6) = -1\). According to the theorem:

• Since 3 is between 7 and -1, there must be at least one point \(c\) in the interval \[-2, 6\] where \(f(c) = 3\).
• The theorem does not guarantee that \(f(0) = 0\) unless 0 is indeed between 7 and -1 and if the function's specific form supports it.
• The theorem does not assert that the function will remain within specific bounds like \([-2, 6]\) throughout the interval. It guarantees intermediate values without restraining the function within an overall range.

By comprehending this perspective of the theorem, you can determine whether the indicated values will be reached within the given interval.